I am working on a problem (problem 10.19 (d)) in John M. Lee's Introduction to Smooth Manifold.
Assume that $\pi$: $E$ $\rightarrow$ $M$ is a fiber bundle with model fiber $F$, I need to prove if $E$ is compact, then so are $M$ and $F$.
Clearly, $M$ is compact and if we assume that $M$ is Hausdorff, then it follows easily from part (c) of this problem.
($\pi$: $E$ $\rightarrow$ $M$ is proper iff $F$ is compact.)
(Edited: I think it suffices to assume that $M$ is $T_1$, i.e., singletons are closed, then it follows from the continuity that the fibers are compact.)
However, I have no idea how to proceed in the general case. Any hints are appreciated.
Since the author chimed in in the comments, let me record the full answer here.
This is still true without separation assumptions on the base. Indeed, by compactness of $M$ there exists a finite cover $\{U_i \subseteq M\}_{i \in I}$ over which $E$ is trivialised. Without loss of generality we may assume that no strict subset of $\{U_i \subseteq M\}$ covers $M$. Then $$Z = M \setminus \bigcup_{j \neq i} U_j \subseteq U_i$$ is a nonempty closed subset over which $E$ is trivial, i.e. $E|_Z \cong F \times Z$. Since $E|_Z \subseteq E$ is closed, it is compact, hence so is $F$ since it is the image of the first projection $F \times Z \to F$. $\square$